Question

Calculate the work done by the brakes of car of mass 1000kg when its speed is reduced from 20 m/s to 10m/s?​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the car (m) = 1000 kg.
• Initial velocity of the car (u) = 20 m/s.
• Final velocity of the car (v) = 10 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

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Firstly, we need to find the Initial Kinetic energy of the car.

Therefore,

$\large{\boxed{\tt K.E_i = \dfrac{1}{2}mu^2}}$

Substituting the values,

$\large{\tt K.E_i = \dfrac{1}{2} \times 1000 \times (20)^2}$

$\large{\tt K.E_i = \dfrac{1}{2} \times 1000 \times 400}$

$\large{\tt K.E_i = \dfrac{1}{\cancel{2}} \times 1000 \times \cancel{400}}$

$\large{\tt K.E_i = 1000 \times 200}$

It becomes,

$\large{\boxed{\tt K.E_i = 2 \times 10^{5} \: J}}$

The initial Kinetic energy is 2 × 10⁵ Joules.

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Now, The final Kinetic energy of the car,

$\large{\boxed{\tt K.E_f = \dfrac{1}{2}mv^2}}$

Substituting the values,

$\large{\tt K.E_f = \dfrac{1}{2} \times 1000 \times (10)^2}$

$\large{\tt K.E_f = \dfrac{1}{2} \times 1000 \times 100}$

$\large{\tt K.E_f = \dfrac{1}{\cancel{2}} \times 1000 \times \cancel{100}}$

$\large{\tt K.E_f = 1000 \times 50}$

It becomes,

$\large{\tt K.E_f = 5 \times 10^{4}}$

Raising the power to 5, Then.

$\large{\boxed{\tt K.E_f = 0.5 \times 10^{5} \: J}}$

The final Kinetic energy is 0.5 × 10⁵ Joules.

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Now, From Work - energy Theorem,

$\large{\boxed{\tt W = K.E_f - K.E_i}}$

Here

• "W" denotes Work done.
• ${\tt K.E_i}$ denotes initial Kinetic energy.
• ${\tt K.E_f}$ denotes Final Kinetic energy.

Now,

Substituting the values,

$\large{\tt W = 0.5 \times 10^5 - 2 \times 10^5}$

$\large{\tt W = (0.5 - 2) \times 10^5}$

It comes as,

$\huge{\boxed{\boxed{\tt W = - 1.5 \times 10^5 \: J}}}$

So, The work done by the brakes to reduce its speed from 20 m/s to 10 m/s is -1.5 × 10⁵ Joules.

Note:-

• Here negative sign indicates that the work done is against the motion of the car.

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Answer 2

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From the Question,

• Mass of the Car,m = 1000 Kg

• Initial Velocity,u = 20 m/s

• Final Velocity,v = 10 m/s

To find

Work Done by the brakes of the car

Work Kinetic Energy Theorem

The rate of work done by a net force is equal to the change in Kinetic Energy

$\huge{\boxed{\boxed{\green{\rm{W = \Delta{K}}}}}}$

${\large\mathtt{\Delta{K} = {K}_{f} - {K}_{i}}}$

Initial Kinetic Energy :

$\sf{{K}_{i} = \frac{1000}{2} \times {(20)}^{2}} \\ \\ \implies \ \sf{{K}_{i} = \frac{{4 \times 10}^{5}}{2} } \\ \\ \implies \ \boxed{\sf{{K}_{i} = 2 \times {10}^{5}}}$

Final Kinetic Energy :

$\displaystyle{\sf{{K}_{f} = \frac{1000}{2} \times {(-10)}^{2}}}\\ \\ \implies \ \boxed{\sf{{K}_{f} = 0.5 \times {10}^{5}}}}$

Putting the values,we get:

$\\ \\ \sf{W = \frac{1}{2}mv {}^{2} - \frac{1}{2} mu {}^{2} } \\ \\ \longrightarrow \: \sf{W = \frac{m}{2}(v {}^{2} - u {}^{2}) } \\ \\ \longrightarrow \: \sf{W = (0.5 \times {10}^{5} - 2 \times {10}^{5})} \\ \\ \huge{ \rm{ \longrightarrow \: W \: = - 1.5 \times {10}^{5} \: J}}$

Thus,the work done by the brakes is -1.5 × 10^5 J

The negative sign indicates force is applied in direction opposite to the relative velocity of the car. Hence, Negative Work